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    <p>If you mean the characters of the numbers (such as 1927 and 9721), there are (at least) a couple of approaches.</p> <p>If you were allowed to sort, one approach is to simply <code>sprintf</code> them to two buffers, sort the characters in the buffers, then see if the strings are equal.</p> <p>However, given your desire to <em>not</em> sort the digits, another alternative is to set up a ten-element array, with all elements initially set to zero, then process each digit in the first number, incrementing the relevant element.</p> <p>Then do the same with the second number but decrementing.</p> <p>If, at the end, it's still all zeros, the numbers were a permutation of each other.</p> <p>This is efficient in that it's an <code>O(n)</code> algorithm where <code>n</code> is the number of digits in the two numbers. The pseudo-code for such a beast would be something like:</p> <pre><code>def arePermutations (num1, num2): create array count, ten elements, all zero. for each digit in num1: increment count[digit] for each digit in num2: decrement count[digit] for each item in count: if item is non-zero: return false return true </code></pre> <p>In C, the following complete program illustrates how this can be done:</p> <pre><code>#include &lt;stdio.h&gt; #include &lt;stdlib.h&gt; #define FALSE (1==0) #define TRUE (1==1) int hasSameDigits (long num1, long num2) { int digits[10]; int i; for (i = 0; i &lt; 10; i++) // Init all counts to zero. digits[i] = 0; while (num1 != 0) { // Process all digits. digits[num1%10]++; // Increment for least significant digit. num1 /= 10; // Get next digit in sequence. } while (num2 != 0) { // Same for num2 except decrement. digits[num2%10]--; num2 /= 10; } for (i = 0; i &lt; 10; i++) if (digits[i] != 0) // Any count different, not a permutation. return FALSE; return TRUE; // All count identical, was a permutation. } </code></pre> <p>&nbsp;</p> <pre><code>int main (int c, char *v[]) { long v1, v2; if (c != 3) { printf ("Usage: %s &lt;number1&gt; &lt;number2&gt;\n", v[0]); return 1; } v1 = atol (v[1]); v2 = atol (v[2]); if (hasSameDigits (v1, v2)) { printf ("%d and %d are permutations\n", v1, v2); } else { printf ("%d and %d are not permutations\n", v1, v2); } return 0; } </code></pre> <p>Simply pass it two (positive) numbers and, assuming they fit in a <code>long</code>, it'll tell you whether they have the same digit counts.</p>
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    1. POChecking whether two numbers are permutation of each other?
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    1. COis it possible to do it without doing any type of sorting?
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    2. COYes, @Ravi, see the update.
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    3. COcan you also suggest something on the line of hashing as well?
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